Question

At an urgent care facility, patients arrive at an average rate of one patient every seven minutes. Assume that the duration between arrivals is exponentially distributed.

a. Find the probability that the time between two successive visits to the urgent care facility is less than $2$minutes.

b. Find the probability that the time between two successive visits to the urgent care facility is more than $15$minutes.

c. If $10$minutes have passed since the last arrival, what is the probability that the next person will arrive within the next five minutes?

d. Find the probability that more than eight patients arrive during a half-hour period.

Short answer

a. $0.2845$

b. $0.1173$

c. $0.4895$

d.$0.0311$

Step-by-step solution

introduction

Probability is just the way that probably something is to occur. Whenever we're uncertain about the result of an occasion, we can discuss the probabilities of specific results — how likely they are. The investigation of occasions represented by likelihood is called insights.

Explanation (part a)

Let duration (in minutes) between successive visits. Since patients arrive at a rate of one patient every seven minutes, and the decay constant is

The cdf is

the probability that the time between two successive visits to the urgent care facility is less than 2 minutes:

$P(T<2)=1-1-e^{-\frac{2}{7}}\approx 0.2485$

Explanation (part b)

the probability that the time between two successive visits to the urgent care facility is more than 15 minutes is

$P(T>15)=1-P(T<15)=1-\left(1-e^{-\frac{15}{7}}\right)\approx e^{-\frac{15}{7}}\approx 0.1173$mins

Explanation (part c)

If 10 minutes have passed since the last arrival, the probability that the next person will arrive within the next five minutes is

$P(T>15\mid T>10)=P(T>5)=1-\left(1-e^{-\frac{5}{7}}\right)=e^{-\frac{5}{7}}\approx 0.4895$mins

Explanation (part d)

Let X= no. of patients arriving during a half-hour period.

Then $X$has the Poisson distribution with a mean of

$\frac{30}{7}$$X~Poisson\left(\frac{30}{7}\right)$

$P(X>8)=1-P(X\le 8)\approx 0.0311$

the probability that more than eight patients arrive during a half-hour period is$0.0311$